Question

Suppose that in a random selection of 100 colored​ candies, 30​% of them are blue. The candy company claims that the percentage of blue candies is equal to 28​%. Use a 0.10 significance level to test that claim.

Identify the test statistic for this hypothesis test.

Identify the​ P-value for this hypothesis test.

Answer 1

Concepts and reason

Statistical hypotheses testing: Hypotheses testing is used to make inferences about the population based on the sample data. The hypotheses test consists of null hypothesis and alternative hypothesis.

Null hypothesis: The null hypothesis states that there is no difference in the test, which is denoted by ${H_0}$ . Moreover, the sign of null hypothesis is equal $\left( = \right)$ , greater than or equal $\left( \ge \right)$ and less than or equal $\left( \le \right)$ .

Alternative hypothesis: The hypothesis that differs from the ${H_0}$ is called alternative hypothesis. This signifies that there is a significant difference in the test. The sign of alternative hypothesis is less than $\left( < \right)$ , greater than $\left( > \right)$ , or not equal $\left( \ne \right)$ .

Proportion: The ratio of the number of favorable outcomes and total number of possible outcomes is in the sample called as the proportion.

Z-statistic for proportion: The standardized z- statistic is defined as the ratio of the ‘distance of the observed statistic from the proportion of the null distribution’ and the ‘standard deviation of the null distribution’.

P-value: The probability of getting the value of the statistic that is as extreme as the observed statistic when the null hypothesis is true is called as P-value. Therefore, it assumes “null hypothesis is true”.

Fundamentals

The formula for test statistic is,

$z = \frac{{\hat p - {p_0}}}{{\sqrt {\frac{{{p_0}\left( {1 - {p_0}} \right)}}{n}} }}$

Here,

$\begin{array}{c}\\n:{\rm{Sample}}\,{\rm{size}}\\\\\hat p:{\rm{Sample}}\,{\rm{proportion}}\\\\{p_0}:{\rm{Population}}\,{\rm{proportion}}\\\end{array}$

The formula for the sample proportion is, $\hat p = \frac{x}{n}$ , here x is the number of success and n is the sample size.

Procedure for finding the probability value (confidence level) from standard normal table:

Let the z value be 0.ab

1.In the standard normal table first locate the value 0.a in the first z column.

2.Locate the value of 0.0b in the first z row.

3.Move right until the column of 0.0b is reached.

4.From the located 0.0b column move down until the row 0.a is reached.

5.Locate the probability value, by the intersection of the row and column values gives the area to the left of z.

The hypotheses are stated below:

From the give information, suppose that a random selection of 100 colored candies out of them 30% are blue. The candy company claims that the percentage of blue candies is equal to 28% and the level of level of significance is 0.10.

That is, $\hat p = 0.30$ and ${p_0} = 0.28$

The hypotheses are,

Null hypothesis:

${H_0}:{p_0} = 0.28$

Alternative hypothesis:

${H_a}:{p_0} \ne 0.28$

The test statistic is obtained as shown below:

From the given information, $n = 100$ , $\hat p = 0.30$ and ${p_0} = 0.28$ .

The test statistic is,

$\begin{array}{c}\\Z = \frac{{\hat p - {p_0}}}{{\sqrt {\frac{{{p_0}\left( {1 - {p_0}} \right)}}{n}} }}\\\\ = \frac{{\left( {0.30 - 0.28} \right)}}{{\sqrt {\frac{{0.28\left( {1 - 0.28} \right)}}{{100}}} }}\\\\ = \frac{{\left( {0.02} \right)}}{{\sqrt {\frac{{0.28\left( {0.72} \right)}}{{100}}} }}\\\\ = \frac{{0.02}}{{\sqrt {0.002} }}\\\end{array}$

$\begin{array}{l}\\ = \frac{{0.02}}{{0.044}}\\\\ = 0.454\\\end{array}$

The p-value is obtained as shown below:

The p-value is,

$\begin{array}{c}\\p - {\rm{value}} = 2\left[ {P\left( {Z > 0.45} \right)} \right]\\\\ = 2\left[ {1 - P\left( {Z \le 0.45} \right)} \right]\\\end{array}$

Procedure for finding the probability value from standard normal table:

The z value be 0.45.

1.In the standard normal table, first locate the value 0.45 in the first z column.

2.Locate the value of 0.4 in the first z row.

3.Move right until the column of 0.05 is reached.

4.From the located 0.05 column move down until the row 0.4 is reached.

5.Locate the probability value, by the intersection of the row and column values gives the area to the left of z.

The area to the left of $Z < 0.45$ is,

$\begin{array}{c}\\P\left( {Z < 0.45} \right) = P\left( {Z < 0.45} \right)\\\\ = 0.6736\\\end{array}$

Now the p- value is,

$\begin{array}{c}\\p - {\rm{value}} = 2\left( {1 - 0.6736} \right)\\\\ = 2\left( {0.3264} \right)\\\\ = 0.6528\\\end{array}$

[Part b]

Ans:

The test statistic for this hypothesis test is 0.45.

Part b

The​ P-value for this hypothesis test is 0.6528.